ta có (x-1)(x-2)(x-3)(x-4)-15=(x-1)(x-4)(x-2)(x-3)-15=\(\left(x^2-5x+4\right)\left(x^2-5x+6\right)-15\)(*)

đặt \(t=x^2-5x+5\)thì pt (*) =\(\left(t-1\right)\left(t+1\right)-15=t^2-1-15\)\(=t^2-16=\left(t+4\right)\left(t-4\right)=\)\(\left(x^2-5x+5+4\right)\left(x^2-5x+5-4\right)=\)\(\left(x^2-5x+9\right)\left(x^2-5x+1\right)\)

um có j đó sai sai

bạn cứ đọc kĩ đi cái đọan  j ẩn phụ ý bạn thử thay t vào là hiểu

con đoạn nhân thì mình lấy cái cuối nhân cái đầu hai cái còn lại nhân vs nhau

\(\dfrac{xy}{2}-x+\dfrac{x^2}{4}=x\left(\dfrac{y}{2}-1+\dfrac{x}{4}\right)\)

x^3-x+6=x^3+2x^2-2x^2-4x+3x+6=x^2.(x+2)-2x.(x+2)+3.(x+2)=(x^2-2x+3).(x+2)

\(4x-x^3+2x^2y-xy^2\)

\(=x\left(4-x^2+2xy-y^2\right)\)

\(=x\left(4-\left(x^2-2xy+y^2\right)\right)\)

\(=x\left(2^2-\left(x-y\right)^2\right)\)

\(=x\left(2+x-y\right)\left(2-x+y\right)\)

\(4x-x^3+2x^2y-xy^2=x\left(4-x^2+2xy-y^2\right)\)

\(=x\left[4-\left(x^2-2xy+y^2\right)\right]=x\left[4-\left(x-y\right)^2\right]=x\left(2-x+y\right)\left(2+x-y\right)\)

\(\left(x-y\right)^3-x^3+y^3=\left(x-y\right)^3-\left(x^3-y^3\right)=\left(x-y\right)^3-\left(x-y\right)\left(x^2+xy+y^2\right)=\left(x-y\right)\left(x^2-2xy+y^2-x^2-xy-y^2\right)=-3xy\left(x-y\right)\)

\(\left(x-y\right)^3-x^3+y^3\\ =\left(x-y\right)^3-\left(x^3-y^3\right)\\ =\left(x-y\right)^3-\left(x-y\right)\left(x^2+xy+y^2\right)\\ =\left(x-y\right)\left[\left(x-y\right)^2-\left(x^2+xy+y^2\right)\right]\\ =\left(x-y\right)\left(x^2-2xy+y^2-x^2-xy-y^2\right)\\ =\left(-3xy\right)\left(x-y\right)\)

\(\left(x-y\right)^3-x^3+y^3\)

\(=x^3-3x^2y+3xy^2-y^3-x^3+y^3\)

\(=-3x^2y+3xy^2\)

\(=-3xy\left(x-y\right)\)

Lỗi?

\(x^2-xy-10x+10y\)

\(=x\left(x-y\right)-10\left(x-y\right)\)

\(=\left(x-y\right)\left(x-10\right)\)

Bài khó quá

4x⁴+4x-3= (2x)²+2(2x)+1-4

=(2x+1)²-2²=(2x+1-2)(2x+1+2)

=(2x-1)(2x+3)

bn làm đ rùi, tui đ cho bn